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Math Help - Arclength of linear curve

  1. #1
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    Arclength of linear curve

    I'm to calculate the line segment y=mx+b between (x0, y0) and (x1, y1), using the formula for arclength when y is a function of x, that is:

    L = (integral evaluated at a and b) (root)(1 + (f'(x))^2

    Which in this case should give:

    L = (integral evaluated at x0 and x1) (root) (1+m^2), as m is the derivative of mx+b.

    This I get to be [(root)(1+m^2)*t]evaluated at x0 and x1, but the answer in my textbook is (1+m^2)|x0-x1|... any help would be appreciated! Thanks!
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    I am not sure about your books answer since m is just a constant.

    \int_{x_0}^{x_1}\sqrt{1+m^2}dx

    \sqrt{1+m^2} this is just some arbitrary number so move outside the integral.

    =\sqrt{1+m^2}*\int_{x_0}^{x_1}dx

    =\sqrt{1+m^2}*\big[x\big]_{x_0}^{x_1}

    =\sqrt{1+m^2}*(x_1-x_0)

    This what I obtained as a solution to the definite integral.
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  3. #3
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    Quote Originally Posted by gralla55 View Post
    I'm to calculate the line segment y=mx+b between (x0, y0) and (x1, y1), using the formula for arclength when y is a function of x, that is:

    L = (integral evaluated at a and b) (root)(1 + (f'(x))^2

    Which in this case should give:

    L = (integral evaluated at x0 and x1) (root) (1+m^2), as m is the derivative of mx+b.

    This I get to be [(root)(1+m^2)*t]evaluated at x0 and x1, but the answer in my textbook is (1+m^2)|x0-x1|... any help would be appreciated! Thanks!
    remember that you're setting up a general case arc length integral.

    what if x_1 < x_0 ?
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    Skeeter the book also got rid of the radical unless he forgot to type it.
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  5. #5
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    Yes, the missing root sign is what confused me... but I guess the book could be wrong. Thanks anyway!
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